I suspect it is quite simple actually but I don't have the knowledge of where I should be looking to solve this one... I thought I might pop this up on hubski in the meantime; I know there is a mathematical bent here. A small part of my work involves simulating sequences of years that are classified as either wet or dry. There is good reason for assuming that this behaviour can be modelled as a two-state markov chain, with different probabilities for transitioning from W>D, D>W etc.
Suppose then I have a sequence of years that I want to use to derive these transition probabilities e.g. D-WWW-DD-WW-DD-W-D-WWWW.
I can calculate the average dry and wet lengths - 1.5 and 2.5 years respectively. I can also estimate (I think) the transition probabilities by taking the number of transitions from a state over the chances that there had been to transition from that state. So assuming the first and last are not intersecting a sequence, I would have for the wet case 4 transitions over 10 wet years, i.e. P = 0.4. The same process for dry gives P = 0.67.
Now for the interesting part - I was wondering whether there was any relationship between the average length and the transition probability. Intuitively it should be the case, as the length will ultimately be determined by the behaviour caused by the probabilities.
Because each iteration behaves like a biased coin toss, I started thinking that the binomial distribution might have something to do with it. An interesting property I noticed is that 1 + the infinite sum of (P^n) from n=1 gives the average lengths as calculated above.
Of course each term of the sum here is the probability of n successess in n tests according to the binomial distribution. As I said, I am sure there is a fairly simple explanation, or perhaps I have gotten something wrong. Either way I suppose I hope someone could shed some light on this, or at least finds this interesting.
Bonus: some nice graphs from the model showing stochastic rainfall and evaporation predictions.